Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, \(0.500 \mathrm{M}\) silver nitrate is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is \(0.641 \mathrm{~g}\). a. Calculate the mass percent of magnesium chloride in the mixture. b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.

When silver nitrate is added to the mixture, it reacts with magnesium nitrate and magnesium chloride separately. Based on solubility rules, silver nitrate will react with magnesium chloride to form silver chloride precipitate. The chemical equations can be written as follows: 1. \(Mg(NO_3)_2 (aq) + 2AgNO_3 (aq) \rightarrow 2AgNO_3 (aq) + Mg(NO_3)_2 (aq)\) (No precipitate forms in this reaction) 2. \(MgCl_2 (aq) + 2AgNO_3 (aq) \rightarrow 2AgCl (s) + Mg(NO_3)_2 (aq)\) (Silver chloride precipitate forms in this reaction) Now that we have the chemical equations, we can proceed to calculate the mass of magnesium chloride in the mixture. Step 2: Calculate moles of the white precipitate

Given the mass of the white precipitate (silver chloride) formed is \(0.641 \mathrm{~g}\). From the periodic table, we can find the molar masses as: Molar mass of silver chloride, \(AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 \mathrm{~g/mol}\) Now we can calculate the moles of silver chloride precipitate formed: moles of \(AgCl = \frac{0.641 \mathrm{~g}}{143.32 \mathrm{~g/mol}} = 0.00447 \mathrm{~mol}\) Step 3: Calculate the mass of magnesium chloride in the mixture

From the balanced chemical equation for the reaction between magnesium chloride and silver nitrate, we can tell that each mole of magnesium chloride reacts with two moles of silver nitrate to form two moles of silver chloride. Using stoichiometry, we can calculate the moles and mass of magnesium chloride in the mixture: moles of \(MgCl_2 = \frac{1}{2} \times\) moles of \(AgCl = \frac{1}{2} \times 0.00447 \mathrm{~mol} = 0.00224 \mathrm{~mol}\) Molar mass of magnesium chloride, \(MgCl_2 = 24.31 (Mg) + 2 \times 35.45 (Cl) = 95.21 \mathrm{~g/mol}\) Mass of magnesium chloride in the mixture: \(mass_{MgCl_2} = moles \times molar \thinspace mass \) \(mass_{MgCl_2} = 0.00224 \mathrm{~mol} \times 95.21 \mathrm{~g/mol} = 0.213 \mathrm{~g}\) Step 4: Calculate the mass percent of magnesium chloride in the mixture

The mass percent of magnesium chloride in the mixture can be calculated as: \(mass \thinspace percent_{MgCl_2} = \frac{mass_{MgCl_2}}{total \thinspace mass \thinspace of \thinspace mixture} \times 100\) \(mass \thinspace percent_{MgCl_2} = \frac{0.213 \mathrm{~g}}{1.50 \mathrm{~g}} \times 100 = 14.2 \% \) So, the mass percent of magnesium chloride in the mixture is 14.2 %. Step 5: Calculate the minimum volume of silver nitrate needed

We have the concentration of silver nitrate solution (\(0.500 \mathrm{M}\)) and the moles of silver chloride precipitate formed (0.00447 mol). From the stoichiometry of the reaction, we know that 1 mole of silver chloride is formed from 1 mole of silver nitrate. So, we need the same moles of silver nitrate (0.00447 mol) to ensure complete formation of the precipitate. Now we can calculate the volume of silver nitrate solution needed: volume = \(\frac{moles}{concentration}\) volume = \(\frac{0.00447 \mathrm{~mol}}{0.500 \mathrm{M}}\) volume = \(0.00894 \mathrm{~L} = 8.94 \mathrm{~mL}\) So, the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate is 8.94 mL.